ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 98

\[\boxed{\text{98.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{4^{\backslash y - 2}}{y + 2} - \frac{3^{\backslash y + 2}}{y - 2} + \frac{12}{y^{2} - 4} =\]

\[= \frac{4y - 8 - 3y - 6 + 12}{y^{2} - 4} =\]

\[= \frac{y - 2}{(y - 2) \cdot (y + 2)} = \frac{1}{y + 2}\]

\[\textbf{б)}\ \frac{a^{\backslash a + 6}}{a - 6} - \frac{3^{\backslash a - 6}}{a + 6} + \frac{a^{2}}{36 - a^{2}} =\]

\[= \frac{a^{2} + 6a - 3a + 18 - a^{2}}{a^{2} - 36} =\]

\[= \frac{3a + 18}{a^{2} - 36} = \frac{3(a + 6)}{(a - 6)(a + 6)} =\]

\[= \frac{3}{a - 6}\]

\[\textbf{в)}\ \frac{x^{2}}{(x - y)^{2}} - \frac{x + y}{2x - 2y} =\]

\[= \frac{{x^{2}}^{\backslash 2}}{(x - y)^{2}} - \frac{x + y^{\backslash x - y}}{2(x - y)} =\]

\[= \frac{2x^{2} - x^{2} + y^{2}}{2{(x - y)}^{2}} = \frac{x^{2} + y^{2}}{2{(x - y)}^{2}}\]

\[\textbf{г)}\ \frac{b}{(a - b)^{2}} - \frac{a + b}{b^{2} - ab} =\]

\[= \frac{b^{\backslash b}}{(a - b)^{2}} - \frac{a + b^{\backslash a - b}}{b \cdot (b - a)} =\]

\[= \frac{b^{2} + a^{2} - b^{2}}{b \cdot (a - b)^{2}} = \frac{a^{2}}{b \cdot (a - b)^{2}}\]